Question

If 3^x+8 = 27^2x+1, then the value of \(\left[\left(\frac{\sqrt{289}}{\sqrt[3]{216}}\right) ^x \div \left(\frac {17}{\sqrt[4]{1296}}\right)^x\right]^{1/2}\) is

(a) 1

(b) 0

(c) \(\frac{17}6\)

(d) \(\frac6{17}\)

Answer 1

Correct option is (a) 1

3^x+8 = 27^2x+1

⇒ 3^x+8 = (3³)^2x+1

⇒ 3^x+8 = 3^6x+3

Base are same so

x + 8 = 6x + 3

x−6x = 3 − 8

⇒ −5x = −5

⇒ x = 1

If x = 1 then the value of

\(\left[\left(\frac{\sqrt{289}}{\sqrt[3]{216}}\right) ^x \div \left(\frac {17}{\sqrt[4]{1296}}\right)^x\right]^{1/2}\)

\(=\left[\frac{\sqrt{(17)^2}}{\sqrt[3]{(6)^3}} \div \frac {17}{\sqrt[4]{(6)^4}}\right]^{1/2}\)

\(= \left[\frac{17}6 \times \frac 6{17}\right] ^{1/2}\)

\(= [1]^{1/2}\)

⇒ \(\sqrt 1 = 1\)